Integration in cylindrical coordinates (r,θ,z){\displaystyle (r,\theta ,z)} is a simple extension of polar coordinates from two to three dimensions. This coordinate system works best when integrating cylinders or cylindrical-like objects. As with spherical coordinates, cylindrical coordinates benefit from lack of dependency between the variables, which allows for easy factoring.

Steps

1

Recall the coordinate conversions. Coordinate conversions exist from Cartesian to cylindrical and from spherical to cylindrical. Below is a list of conversions from Cartesian to cylindrical. Above is a diagram with point P{\displaystyle P} described in cylindrical coordinates.

Set up the coordinate-independent integral. We are dealing with volume integrals in three dimensions, so we will use a volume differential dV{\displaystyle {\mathrm {d} }V} and integrate over a volume V.{\displaystyle V.}

∫VdV{\displaystyle \int _{V}{\mathrm {d} }V}

Most of the time, you will have an expression in the integrand. If so, make sure that it is in cylindrical coordinates.

Those familiar with polar coordinates will understand that the area element dA=rdrdθ.{\displaystyle {\mathrm {d} }A=r{\mathrm {d} }r{\mathrm {d} }\theta .} This extra r stems from the fact that the side of the differential polar rectangle facing the angle has a side length of rdθ{\displaystyle r{\mathrm {d} }\theta } to scale to units of distance.

4

Set up the boundaries. Choose a coordinate system that allows for the easiest integration.

As with polar coordinates, the range of θ{\displaystyle \theta } is [0,2π],{\displaystyle [0,2\pi ],} unless there are applications to integrating over more than the whole object.

5

Integrate. Once everything is set up in cylindrical coordinates, simply integrate using any means possible and evaluate.

To save space in this article (and in your calculations) for the moment of inertia of a cone, it is useful to recognize the integral ∫02πsin2⁡θdθ=π.{\displaystyle \int _{0}^{2\pi }\sin ^{2}\theta {\mathrm {d} }\theta =\pi .}

Method1

Volume of a Cylinder

1

Calculate the volume of a cylinder of radius R and height h.

Choose a coordinate system such that the radial center of the cylinder rests on the z-axis. The bottom of the cylinder will be on the z=0{\displaystyle z=0} plane for simplicity of calculations.

Note that we could have swapped the integrals. The end result would be the same. However, in more general cases, the boundaries will not stay the same, so the order in which you integrate does matter.

Method2

Moment of Inertia of a Cone

1

Calculate the moment of inertia of a right circular cone. This cone is centered on the z-axis with the apex at the origin, but rotates with respect to the x-axis. In other words, it is rotating laterally, similar to how a beam from a lighthouse rotates. Assume this cone has a height h,{\displaystyle h,} radius a,{\displaystyle a,} mass M,{\displaystyle M,} and constant density σ.{\displaystyle \sigma .}

Most moment of inertia questions are written with answers in terms of M{\displaystyle M} and R{\displaystyle R} (in this example, a{\displaystyle a}), but because a cone also requires a specified height, there will be a term with h{\displaystyle h} in it as well.

2

Recall the moment of inertia formula.

I=∫Mr2dm,{\displaystyle I=\int _{M}r^{2}{\mathrm {d} }m,} where r=y2+z2{\displaystyle r={\sqrt {y^{2}+z^{2}}}} is the perpendicular distance from the axis (the cone is rotating about the x-axis) and we are integrating over the mass M.{\displaystyle M.}

3

Recall the relationship between mass, volume, and density when density is constant.

σ=MV.{\displaystyle \sigma ={\frac {M}{V}}.} Of course, we know the volume of the cone as 13πa2h,{\displaystyle {\frac {1}{3}}\pi a^{2}h,} so

σ=3Mπa2h.{\displaystyle \sigma ={\frac {3M}{\pi a^{2}h}}.}

4

Obtain the boundaries. We face a dilemma here - we are not integrating over a cylinder, but a cone. Instead, notice the relationships between the variables of integration. As z{\displaystyle z} increases, r{\displaystyle r} increases as well. Therefore, there is variable dependency in the integration, and one of the boundaries will no longer be a constant.

Solve for either the radius or height. Both cases are completely equivalent, but be careful of the boundaries that result, for they are not the same. We will solve for radius and compute the integral that results. See the tips for computing the integral after solving for height.

r=azh{\displaystyle r={\frac {az}{h}}}

Then, z{\displaystyle z} integrates from 0{\displaystyle 0} to h,{\displaystyle h,} and r{\displaystyle r} goes from 0{\displaystyle 0} to azh.{\displaystyle {\frac {az}{h}}.} Notice that the nature of the object being integrated over introduces variable dependency in the boundaries. In this case, after we integrate the height, the upper boundary of the radius integral is dependent on the z{\displaystyle z} variable.

5

Rewrite the moment of inertia integral in terms of a volume integral, then solve. The order of the integrals matters here, owing to the way we calculated our boundaries. Also note constants that factor out.

Notice that although cylindrical coordinates do not have as much variable dependency in the integrand as do Cartesian coordinates, that does not mean the dependency goes away. Similar to Cartesian integrals, we will have to manually integrate one at a time.

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Tips

In the moment of inertia example, we solved the equation of the cone for r.{\displaystyle r.} What if we solved for z?{\displaystyle z?}

z=hra{\displaystyle z={\frac {hr}{a}}}

Then, r{\displaystyle r} would go from 0{\displaystyle 0} to a,{\displaystyle a,} but z{\displaystyle z} would go from hra{\displaystyle {\frac {hr}{a}}} to h.{\displaystyle h.} The resulting integral would be like so.

This is not a very intuitive result, for it is the lower boundary that ends up having the r{\displaystyle r} dependency. One way to think about this is to understand that when you integrate the radius integral first, you are ignoring whatever happens on the z-axis. After you sum from 0{\displaystyle 0} to a,{\displaystyle a,} there is a region in which you cannot integrate over, because there is only empty space there. We account for that fact when we set the side of the cone as the lower boundary.

Similar to the above example, you are first integrating height, but that integral has no information on what happens radially. Since the radius starts with 0 and increases with height, it is easy to set the boundaries.

If you are still unconvinced, solve the integral. You will get the same answer either way.

Solve the moment of inertia problem with respect to the y-axis. You will find that your answer is equivalent, because the density of the cone is constant. An easy way to recognize why your answer is the same is if you simply rotated the xy-plane.

Do not use the variable of integration in the integral boundary, as this makes the process highly unclear and may lead to confusion. Take extra care to distinguish your variables by using uppercase/lowercase letters, tildes ('squiggly' signs) above the variable of integration, or even different letters.